Solving Fixed Point Problems: x=4-x^2, f(x)=7+sqrt(x-1), f(x)=sqrt(10+3x)-4

[Painguy]

Homework Statement

Find all real values x that are fixed by the function y=4-x^2
f(x)=4-x^2

Homework Equations

x=y

The Attempt at a Solution

x=4-x62
0=-x^2-x+4
0=-(x^2+x+(1/4))+(17/4)

This is where i get stuck.

I also have two other problems which iIdo not understand how to work with.
f(x)=7+sqrt(x-1)
f(x)=sqrt(10+3x) -4

The main problem keeping me from doing the the two above is not knowing what to do with the square root of the expression underneath. Thanks in advance.

 

[verty]

Remember that you want to isolate x. Focus on doing that. If you are completing the square, be sure you know that method.

 

[HallsofIvy]

A "fixed point" for a function f is a value of x such that f(x)= x.

1) $4- x^2= x$ gives $x^2+ x- 4= 0$.
Solve that by completing the square or using the quadratic formula.

2) $7+ \sqrt{x- 1}= x$ is the same as $\sqrt{x- 1}= x- 7$.
Square both sides to get $x- 1= (x- 7)^2= x^2- 14x+ 49$.
That is also a quadratic equation- but this time is easily factorable. Be sure to check your answers in the original equation. "Squaring both sides" of an equation can introduce spurious solutions.

3). $\sqrt{10+ 3x}- 4= x$ is the same as $\sqrt{10+ 3x}= x+ 4$. Again, square both sides to get a quadratic equation. Be sure to check your answers in the original equation.

 

[Painguy]

A "fixed point" for a function f is a value of x such that f(x)= x.

1) $4- x^2= x$ gives $x^2+ x- 4= 0$.
Solve that by completing the square or using the quadratic formula.

2) $7+ \sqrt{x- 1}= x$ is the same as $\sqrt{x- 1}= x- 7$.
Square both sides to get $x- 1= (x- 7)^2= x^2- 14x+ 49$.
That is also a quadratic equation- but this time is easily factorable. Be sure to check your answers in the original equation. "Squaring both sides" of an equation can introduce spurious solutions.

3). $\sqrt{10+ 3x}- 4= x$ is the same as $\sqrt{10+ 3x}= x+ 4$. Again, square both sides to get a quadratic equation. Be sure to check your answers in the original equation.

So for the 1st problem would the answer be $(\sqrt{17}/2)-(1/2)$ or $(-\sqrt{17}/2)-(1/2)$ Thanks for all of your help by the way.